Can someone help?? I don’t understand this at all

The segments shown below could form a triangle.<br><br> A. True<br><br> B. False

muminat

Answer:

the answer is true :A

5 0

4 years ago

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g A probability experiment is conducted in which the sample space of the experiment is Upper S equals StartSet 9 comma 10 comma

Mars2501 [29]

Answer:

$P(E^c)=\dfrac{1}{3}$

Step-by-step explanation:

The sample space for the experiment is:

$S=\{9,10,11,12,13,14,15,16,17,18,19,20\}$

Number of elements in S, n(S)=12

Given an event E such that:

$E=\{10,11,12,13,14,15,16,17\}$

Number of elements in E, n(E)=8

The outcomes in the complement of E are the outcomes in S that are not in E.

$E^c=\{9,18,19,20\}$

Number of elements in the complement of E, $n(E^c)=4$

Therefore:

$P(E^c)=\dfrac{n(E^c)}{n(S)}\\ =\dfrac{4}{12}\\P(E^c)=\dfrac{1}{3}$

8 0

3 years ago

Which of the following statements best describes the effect of replacing the graph of y = f(x) with the graph of y = f(x + 6)? T

Alenkasestr [34]

Answer:

The graph of y = f(x) will shift left 6 units, as shown in figure a.

Step-by-step explanation:

A translation means we are able to move the graph of a function up or down - normally called Vertical Translation - and right or left - commonly called Horizontal Translation.

If we replace the graph of y = f(x) with the graph of y = f(x + 6). It means we have added 6 units to the input, meaning the graph y = f(x) will shift left by 6 units as 6 is being added directly to the x, so it is f(x + 6).

For example, if y = x² is the original function and of 6 is directly added to x i.e. f(x + 6), making it y = (x + 6)². So, we can easily observe that there will be a horizontal translation left of 8 units, as shown in figure a.

So, the graph of y = f(x) will shift left 6 units.

Keywords: graph shift, translation

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5 0

4 years ago

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A subset $S \subseteq \mathbb{R}$ is called open if for every $x \in S$, there exists a real number $\epsilon > 0$ such that

const2013 [10]

Answer:

Step-by-step explanation:

REcall that given sets S,T if we want to prove that $S\subseteqT$ , then we need to prove that for all x that is in S, it is in T.

a) Let (a,b) be a non empty interval and $x\in (a,b)$ . Then a<x <b. Let $\varepsilon = \min{\min\{b-x, x-a\}}{2}$ Consider $y \in (x-\varepsilon,x+\varepsilon)$ , then

$y$ and

$y>x-\varepsilon>x-(x-a) = a$ .

Then $y\in (a,b)$ . Hence, (a,b) is open.

Consider the complement of [a,b] (i.e $(a,b)^c$ ).

Then, it is beyond the scope of this answer that

$(a,b)^c = (-\infty,a) \cup (b,\infty)$ .

Suppose that $x\in (a,b)^c$ and without loss of generality, suppose that x < a (The same technique applies when x>b). Take $\varepsilon = \frac{a-x}{2}$ and consider $y \in (x-\varepsilon,x+\varepsilon)$ . Then

$y$

Then y \in (-\infty,a). Applying the same argument when $x \in (b,\infty)$ we find that [a,b] is closed.

c) Let I be an arbitrary set of indexes and consider the family of open sets $\{A_i\}_{i\in I}. Let [tex]B = \bigcup_{i\in I}A_i$ . Let $x \in B$ . Then, by detinition there exists an index $i_0$ such that $x\in A_{i_0}$ . Since $A_{i_0}$ is open, there exists a positive epsilon such that $(x-\varepsilon,x+\varepsilon)\subseteq A_{i_0} \subseteq B$ . Hence, B is open.

d). Consider the following family of open intervals $A_n = (a-\frac{1}{n},b+\frac{1}{n})$ . Let $B = \bigcap_{n=1}^{\infty}A_n$ . It can be easily proven that

$B =[a,b]$ . Then, the intersection of open intervals doesn't need to be an open interval.

b) Note that for every $x \in \mathbb{R}$ and for every $\varepsilon>0$ we have that $(x-\varepsilon,x+\varepsilon)\subseteq \mathbb{R}$ . This means that $\mathbb{R}$ is open, and by definition, $\emptyset$ is closed.

Note that the definition of an open set is the following:

if for every $x \in S$ , there exists a real number $\epsilon > 0$ such that $(x-\epsilon,x \epsilon) \subseteq S$ . This means that if a set is not open, there exists an element x in the set S such that for a especific value of epsilon, the subset (x-epsilon, x + epsilon) is not a proper subset of S. Suppose that S is the empty set, and suppose that S is not open. This would imply, by the definition, that there exists an element in S that contradicts the definition of an open set. But, since S is the empty set, it is a contradiction that it has an element. Hence, it must be true that S (i.e the empty set) is open. Hence $\mathbb{R}$ is also closed, by definition. If you want to prove that this are the only sets that satisfy this property, you must prove that $\mathbb{R}$ is a connected set (this is a topic in topology)

6 0

3 years ago

Question 4) Identify the surface area of the composite figure.

nignag [31]

The surface area of the composite figure is 714 sq. in.

<h3>What is formula of surface area of cube ?</h3>

Let, a be the edge of the cube.

Then, surface area of the cube = 6a² square unit

<h3>What is formula of surface area of cuboid ?</h3>

Let, length of the cuboid = l unit

Width of the cuboid = w unit

Height of the cuboid = h unit

Then, surface area of the cuboid = 2(lw+lh+hw) square unit

<h3>What is the required surface area ?</h3>

Given, edge of cube = 4 in.

∴ Surface area of cube = (6×4²) sq. in. = 96 sq. in.

Now, given, length of cuboid = 15 in.

Width of the cuboid = 10 in. & height of the cuboid = 7 in.

∴ Surface area of the cuboid = 2{(15×10)+(15×7)+(10×7)} sq. in.

= 2{150+105+70} sq. in.

= 2{325} sq. in. = 650 sq. in.

Here, we have to find the surface area of the composite figure.

We have to subtract the surface area from both of cube & cuboid, where they are attached.In that case, we have to subtract (4×4) from both.

So, we have to subtract 2(4×4) sq. in. = 32 sq. in.

∴ The surface area of the composite figure = (650+96-32) sq. in.

= 714 sq. in.

Learn more about composite figure here :

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5 0

2 years ago